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Probability of drawing a full house

  1. Jan 28, 2008 #1
    1. The problem statement, all variables and given/known data
    Whats the probability of drawing a full house(three of a kind and two of a kind) from a deck of 52 cards without replacement.


    2. Relevant equations



    3. The attempt at a solution
    I know that the first two cards are fine but next one should either be of the first one drawn or the second one drawn. Then the last two should be of the other kind that wasnt matched previously but I dont know how to look at that in terms of probability. How should I go about doing such a question. Got this in my Mech Eng Lab I class.
     
  2. jcsd
  3. Jan 28, 2008 #2

    Dick

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    There's an easier way to count them. Pick a card type for the three of a kind. There are 13 ways to do that. Now how many ways to pick three cards for each card type? There are 12 choices for the second card type. Now how many ways to pick two cards from each card type? Multiply those all together and you have the total number of possible full house hands. Now divide by the total number of possible hands.
     
  4. Jan 28, 2008 #3
    I understand the 13 and 12 part as there are 13 sets in the 52 card pack and if you pick one kind, you'll clearly have 12 sets that your are able to pick from the second time around. But I am still at a lost. I really don't know how many ways to pick three cards for each card type as well as the latter. *sighs*
     
  5. Jan 28, 2008 #4

    Dick

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    Ok. How many ways to pick 3 is easy. Of the four cards you can leave any one out. There are four ways to do that. This is often written C(4,3)=4. Look up 'combinatorial coefficient'. Similarly the number of ways to choose 2 from 4 is C(4,2)=6. C(n,k) is n!/(k!*(n-k)!). Does that seem familiar? You'll need the general formula to figure out how many ways to choose 5 cards from 52. It's C(52,5). Can you work it out?
     
  6. Jan 28, 2008 #5
    yeah all those formulas have been used by me at least once between college and primary school(junior high i guess for you guys). So now i understand how many ways to pick 3 cards and as a result how to do two as well. Now the result of say 4nCr3. What do i do with it. You said multiply it by 4nCr2. Why was that done? Then you said put it over the total number of possible hands. What would that be? the total number of drawing five cards or something....? Forgive the ignorance please.
     
  7. Jan 28, 2008 #6

    Dick

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    13*C(4,3)*12*C(4,2)/C(52,5) is the answer. Can you tell me why that works in words? I'll start you off. C(52,5) is the total number of possible hands (52 cards, choose 5).
     
  8. Jan 28, 2008 #7
    Oh yes. High school probability is coming back to me. I knew that last part. From you told me earlier that total it should be divided by the total possible outcomes, I figured it would be 52nCr5 as and that will give you the total combination of five cards in general, whether it be full house or not. As for the mulitiplying them, I am not sure why you did that. I remember that we had used this rule that AND was mulitiply and OR was addition. The former applies here cause they're independent if you adjust the cards remaning accordingly. I just need to iron out why it was required to multiply the 13 and 12 in. What does that rule state?
     
  9. Jan 28, 2008 #8

    Dick

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    First I pick 1 type out of 13 AND then I pick one type out of the remaining 12. I have to do both. The same for the number of cards.
     
    Last edited: Jan 28, 2008
  10. Jan 29, 2008 #9
    No no I know why you multiplied each other because they're independent events. But I got it nevertheless after reflecting on the matter. You multiplied the first one by 13 because after finding the ways of drawing three cards from four cards(a set) you needed to find how many would come from 13 sets which would require the multiplication and so and so forth for the other.

    That though brings me to another question.

    Say I am doing a multiple choice paper of 5 questions and I ask whats the probability of guessing the first 3 correctly. Since the probability of getting a question right is 1/5, would the probability of guessing three right by (1/5)^3 = 1/125? Or would i have to use the formula P(n) = (N!/((N-n)!N!))*(P^n)*(1-P)^(N-n)?
     
    Last edited: Jan 29, 2008
  11. Jan 29, 2008 #10

    Dick

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    If the question asks for only the first three correctly (and doesn't care about the other two). Then the first approach is fine. If it asks for the probability of guessing EXACTLY three correctly (and doesn't care which ones) then the second approach is fine.
     
  12. Jan 30, 2008 #11
    Okay thanks man. You have been a big help.
     
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